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 Multiple Choice QuestionsMultiple Choice Questions

21.

A pulley of radius 2m is rotated about its axis by a force F = (20t - 5t2) Newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg-m2, the number of rotations made by the pulley before its direction of motion if reversed is

  • more than 3 but less than 6

  • more than 6 but less than 9

  • more than 9

  • more than 9


A.

more than 3 but less than 6

To reverse the direction
integral τdθ space equals 0
straight tau space equals space left parenthesis 20 space straight t minus 5 straight t squared right parenthesis 2 space equals space 40 straight t minus 10 straight t squared
straight alpha space equals space straight tau over straight I space equals space fraction numerator 40 straight t minus 10 straight t squared over denominator 10 end fraction space equals space 4 straight t minus straight t squared
straight omega space equals space integral subscript 0 superscript straight t space αdt space equals space 2 straight t squared minus straight t cubed over 3
straight omega space is space zero space at
2 straight t squared minus straight t cubed over 3 space equals space 0
straight t cubed space equals space 6 straight t squared
straight t space equals space 6 space sec
straight theta space equals space integral ωdt
space equals space integral subscript 0 superscript 6 left parenthesis 2 straight t squared minus straight t cubed over 3 right parenthesis dt
open square brackets fraction numerator 2 straight t cubed over denominator 3 end fraction minus straight t to the power of 4 over 12 close square brackets subscript 0 superscript 6 space equals space 216 space open square brackets 2 over 3 minus 1 half close square brackets space equals space 36 space rad.
No space of space revolution space fraction numerator 36 over denominator 2 straight pi end fraction space Less space than space 6

869 Views

22.

A mass m hangs with the help of a string wrapped around a pulley on a frictionless bearing. The pulley has mass m and radius R.Assuming pulley to be a perfect uniform circular disc, the acceleration of the mass m, if the string does not slip on the pulley, is

  • g

  • 2/3g

  • g/3

  • g/3


B.

2/3g


mg - T = ma
TR space equals space fraction numerator mR squared straight alpha over denominator 2 end fraction
straight T space equals space mRα over 2 space equals space ma over 2
mg space minus ma over 2 space equals space ma
fraction numerator 3 space ma over denominator 2 end fraction space equals space mg
straight a space equals space fraction numerator 2 straight g over denominator 3 end fraction
514 Views

23.

A body of mass m = 3.513 kg is moving along the x-axis with a speed of 5.00 ms−1.The magnitude of its momentum is recorded as

  • 17.6 kg ms−1 

  • 17.565 kg ms−1

  • 17.56 kg ms−1

  • 17.56 kg ms−1


A.

17.6 kg ms−1 

P = mv = 3.513 × 5.00 ≈ 17.6

184 Views

24.

For a particle in uniform circular motion, the acceleration an at straight a with rightwards arrow on top point P (R, θ) on the circle of radius R is (Here θ is measured from the x–axis)

  • negative straight v over straight R space cos space straight theta space bold i with bold hat on top space plus space straight v squared over straight R space sin space straight theta space bold j with bold hat on top
  • negative straight v over straight R space sin space straight theta space bold i with bold hat on top space plus space straight v squared over straight R space cos space straight theta space bold j with bold hat on top
  • negative straight v over straight R space cos space straight theta space bold i with bold hat on top space minus space straight v squared over straight R space sin space straight theta space bold j with bold hat on top
  • negative straight v over straight R space cos space straight theta space bold i with bold hat on top space minus space straight v squared over straight R space sin space straight theta space bold j with bold hat on top

C.

negative straight v over straight R space cos space straight theta space bold i with bold hat on top space minus space straight v squared over straight R space sin space straight theta space bold j with bold hat on top

straight a with rightwards arrow on top space equals space minus space straight V squared over straight R space cos space straight theta space straight i with hat on top space minus space straight V squared over straight R space sin space straight theta space straight j with hat on top

675 Views

25.

Two fixed frictionless inclined planes making an angle 30° and 60° with the vertical are shown in the figure. Two blocks A and B are placed on the two planes. What is the relative vertical acceleration of A with respect to B?

  • 4.9 ms-2 in horizontal direction

  • 9.8 ms-2 in vertical direction

  • zero

  • zero


D.

zero

For the motion of block along inclined plane
mg sin θ =ma
a = g sin θ
where a is along the inclined plane.
The vertical component of acceleration is g sin2θ
Therefore, the relative vertical acceleration of  A with respect to B is 
g (sin260 - sin230) = g/2 = 4.9 ms-2 (in vertical direction)

453 Views

26.

A liquid in a beaker has temperature θ(t) at time t and θ0 is temperature of surroundings, then according to Newton's law of cooling the correct graph between loge (θ – θ0) and t is


A.

According to Newton's law of cooling.

dθ over dt space proportional to left parenthesis straight theta minus straight theta subscript straight o right parenthesis
rightwards double arrow fraction numerator begin display style dθ end style over denominator begin display style dt end style end fraction space equals space minus straight k left parenthesis straight theta minus straight theta subscript straight o right parenthesis
integral fraction numerator begin display style dθ end style over denominator begin display style straight theta minus straight theta subscript straight o end style end fraction space equals space integral negative kdt space
rightwards double arrow space In space left parenthesis straight theta minus straight theta subscript straight o right parenthesis space equals space minus kt plus straight c

Hence the plot of ln(θ – θ0) vs t should be a straight line with negative slope.

473 Views

27.

A block of mass 0.50 kg is moving with a speed of 2.00 m/s on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

  • 0.16 J

  • 1.00 J

  • 0.67 J

  • 0.67 J


C.

0.67 J

m1u1 + m2u2 = (m1 + m2)v
v = 2/3 m/s
Energy space loss space equals space 1 half space left parenthesis 0.5 right parenthesis space straight x space left parenthesis 2 right parenthesis squared space minus space 1 half space left parenthesis 1.5 right parenthesis space straight x space open parentheses 2 over 3 close parentheses squared
space equals space 0.67 space straight J

183 Views

28.

STATEMENT – 1
Two particles moving in the same direction do not lose all their energy in a completely inelastic collision.
STATEMENT – 2
Principle of conservation of momentum holds true for all kinds of collisions.

  • The statement I is True, Statement II is False.

  • The statement I is True, Statement II is True; Statement II is a correct explanation for Statement I.

  • The statement I is True, Statement II is True; Statement II is not the correct explanation for Statement I.

  • The statement I is True, Statement II is True; Statement II is not the correct explanation for Statement I.


B.

The statement I is True, Statement II is True; Statement II is a correct explanation for Statement I.

264 Views

29.

A particle is projected at 60° to the horizontal with a kinetic energy K. The kinetic energy at the highest point is 

  • K

  • 0

  • K/2

  • K/2


D.

K/2

132 Views

30.

A thin liquid film formed between a U-shaped wire and a light slider supports a weight of 1.5 x10–2N (see figure). The length of the slider is 30 cm and its weight negligible. The surface tension of the liquid film is

  • 0.0125 Nm-1

  • 0.1 Nm-1

  • 0.05 Nm-1

  • 0.05 Nm-1


D.

0.05 Nm-1



The force of surface tension acting on the slider balances the force due to the weight.

⇒F = 2Tl = w
⇒2T(0.3) = 1.5 x 10–2
⇒T = 2.5 x 10–2 N/m

369 Views